If ∫₀ˣf(t)dt = x² + ∫₁ˣ t²f(t)dt, then f'(1/2) is:

∫₀ˣf(t)dt = x² + ∫₁ˣ t²f(t)dt
f'(1/2) = ?
Differentiate w.r.t. 'x'
f(x) = 2x + 0 - x²f(x)
f(x) = 2x/(1 + x²)
⇒f'(x) = (1 + x²)2 - x(2x)(1 + x²)²/ (1 + x²)⁴
f'(1/2) = (1 + (1/2)²) / (1 + (1/2)²)² = (5/4) / (25/16) = 4/5